Answer:
A
Explanation:
To calculate questions like this, you need to make use of the equation below:
E = hν
E is the energy, h is Planck's constant, and ν is the frequency.
Also, another important formula you need for this question is the following:
ν = c / λ
ν is the frequency, c is the speed of light, and λ is the wavelength.
Therefore, E = hν = hc/λ.
Also remember to convert nm wavelength to meters.
To answer the question, E = hc/λ = ((6.626*10^-34 J•s)(3*10^8 m/s))/(4.45*10-7 m) = 4.42 • 10^-19 J = A
Answer:
d. Copper (II) sulfate
Explanation:
Given data:
Mass of Al = 1.25 g
Mass of CuSO₄ = 3.28 g
What is limiting reactant = ?
Solution:
Chemical equation:
2Al + 3CuSO₄ → Al₂ (SO₄)₃ + 3Cu
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 1.25 g/ 27 g/mol
Number of moles = 0.05 mol
Number of moles of CuSO₄:
Number of moles = mass/molar mass
Number of moles = 3.28 g/ 159.6 g/mol
Number of moles = 0.02 mol
now we will compare the moles of reactant with product.
Al : Al₂ (SO₄)₃
2 : 1
0.05 : 1/2×0.05=0.025 mol
Al : Cu
2 : 3
0.05 : 3/2×0.05 = 0.075 mol
CuSO₄ : Al₂ (SO₄)₃
3 : 1
0.02 : 1/3×0.02=0.007 mol
CuSO₄ : Cu
3 : 3
0.02 : 0.02
Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.
Answer:
0.00032 Grams of NaCl per 1 gram of the solution
Explanation:
Answer:3.6 x 101 or 8.77 x 10-1
